Question: A secant line intersects the curve $y=2x^2+1$ at two points with $x$ -coordinates $4$ and $4+h$, where $h\neq0$. What is the slope of the secant line in terms of $h$ ? Your answer must be fully expanded and simplified.
Answer: We are given that the secant line intersects the curve at $x=4$ and $x=4+h$. Since these points are on the curve $y=2x^2+1$, we know that their $y$ -values are $y=33$ and $y=2(4+h)^2+1$, correspondingly. To summarize this part, we know that the secant line passes through the points $(4,33)$ and $(4+h,2(4+h)^2+1)$. This should be enough to find the slope of that line. $\begin{aligned} \text{Slope}&=\dfrac{\text{Change in }y}{\text{Change in }x} \\\\ &=\dfrac{2(4+h)^2+1-33}{4+h-4} \\\\ &=\dfrac{2(4+h)^2-32}{h} \end{aligned}$ We can now simplify the expression we obtained. $\begin{aligned} &\phantom{=}\dfrac{2(4+h)^2-32}{h} \\\\ &=\dfrac{2(16+8h+h^2)-32}{h} \\\\ &=\dfrac{32+16h+2h^2-32}{h} \\\\ &=\dfrac{2h^2+16h}{h} \\\\ &=\dfrac{h(2h+16)}{h} \\\\ &=2h+16\text{, for }h\neq 0 \end{aligned}$ Since we are given that $h\neq 0$, we can conclude that the slope of the secant line is $2h+16$.